We have \(\dfrac{SP}{PB}=\dfrac{area\left(APS\right)}{area\left(ABP\right)}=\dfrac{5}{6}\)

Call the area of PSR be x, the area of CSR be y, we have:

  \(\dfrac{area\left(PSR\right)}{area\left(PBR\right)}=\dfrac{SP}{PB}=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{x}{7}=\dfrac{5}{6}\) \(\Rightarrow x=\dfrac{35}{6}\)

\(\dfrac{BR}{CR}=\dfrac{area\left(BSR\right)}{area\left(CRS\right)}=\dfrac{7+x}{y}\)     (1)

\(\dfrac{BR}{CR}=\dfrac{area\left(ABR\right)}{area\left(ACR\right)}=\dfrac{13}{x+y+5}\)   (2)

(1), (2) => \(\dfrac{7+x}{y}=\dfrac{13}{x+y+5}\)

\(\Rightarrow y=\dfrac{\left(7+x\right)\left(5+x\right)}{6-x}=\dfrac{5005}{6}\)

So, \(area\left(ABC\right)=5+6+7+x+y\)

                             \(=5+6+7+\dfrac{35}{6}+\dfrac{5005}{6}=858\)

A B C R S P 5 6 7 M N x y

We have SPPB=area(APS)area(ABP)=56

Call the area of PSR be x, the area of CSR be y, we have:

  area(PSR)area(PBR)=SPPB=56

⇒x7=56

 ⇒x=356

BRCR=area(BSR)area(CRS)=7+xy

     (1)

BRCR=area(ABR)area(ACR)=13x+y+5

   (2)

(1), (2) => 7+xy=13x+y+5

⇒y=(7+x)(5+x)6−x=50056

So, area(ABC)=5+6+7+x+y

                             =5+6+7+356+50056=858